General Topology [1 ed.]

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MATHEMATICAL

EXPOSITIONS

EDITORIAL

BOARD

S. Beatry, H. S. M. Coxeter, B. A. Grirritx, A. Rosrnson, G. DE B. Rosryson (Secretary),

with the co-operation

G. G. LORENTE, W. J. WEBBER

of

A. A. ALBERT, G. BrrxHoFF, R. Braver, R. L. JEFFERY, C. G. LATIMER, E. J. McSuane, G. Patt, H. P. RoBERTSON, J. L. Synce, G. SzEco, A. W. Tucxer, O. ZARISKI

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L. JEFFERY

SrerPinsxk1 (Translated by C. Ceciu1a Knriecer) Press

No. 8. Bernstein Polynomials—G.G. Lorentz

MATHEMATICAL

EXPOSITIONS,

GENERAL

No. 7

TOPOLOGY

by WACLAW

SIERPINSKI

Professor of Mathematics at the University of Warsaw, Member of the Polish Academy of Arts and Sciences, Corresponding Member of the Institute of France

Translated by C. CECILIA Assistant

UNIVERSITY

KRIEGER

Professor in Mathematics University of Toronto -

OF TORONTO

TORONTO,

1952

PRESS

GEOFFREY OxForD

LONDON: CUMBERLEGE UNIVERSITY

PRESS

CopyRIGHT, CaNnapA, 1952 PRINTED IN CANADA

AUTHOR’S

alice

PREFACE

theorems of any geometry (e.g., Euclidean)

follow, as is well known,

from a number of axioms, i.e., hypotheses about the space considered, and from accepted definitions. A given theorem may be a consequence of some of the axioms and may not require all of them. Such a theorem will be true not only in the space defined by all the axioms, but also in more general spaces. It will, therefore, be of importance to introduce axioms gradually and to deduce from them as many conclusions as possible. We

thus

obtained

$ 3 5

arrive

at the

for a given

satisfies

the

axioms

axioms.

Herein

For,

with

from

that

space

to various

types

abstract of that

lies the

a suitable may

concept

space space;

practical

choice

are

true

however,

to the theory

for each of the

for such

to different

space

the

advantage

of axioms

be applied

of geometry,

of an abstract

set

(Fréchet).

set of elements may

study

a space, branches

of functions,

Theorems

also

which

satisfy

of abstract

the theorems of mathematics,

other spaces.

obtained e-g.,

etc.

In the first chapter we develop a fairly detailed theory of the so-called Fréchet (V)spaces. A Fréchet (V)space is a set K whose elements are subject to only one condition, namely, that with each element p of K there is associated at least one subset of K called a neighbourhood of the element #. In chapter II we investigate (V)spaces which satisfy additional axioms, i.e., the so-called topological spaces; in chapters III, IV, and V we study topological spaces satisfying various additional axioms. Chapter VI is devoted to the study of very important particular topological spaces, namely, the socalled metric spaces, which find numerous applications, and chapter VII deals with the so-called complete metric spaces. It may be said about chapters I, II, V, VI, and VII that in each of them new axioms are introduced about the space under consideration and theorems are derived from them. In general, the theorems of each of these chapters are not true in a space satisfying only the axioms of the preceding chapters.

Such an axiomatic treatment of the theory of point sets, apart from its logical simplicity, has also an advantage in that it supplies excellent material for exercise in abstract thinking and logical argument in the deduction of theorems from stated suppositions alone; i.e., in proving the theorems by drawing logical conclusions only and without any appeal to intuition, which is so apt to mislead one in the theory of sets. Vv

231095

vi

PREFACE

The book differs to quite an extent from the Introduction to General Topology (Toronto, 1934). Apart from a different axiomatic treatment, which seems to us much more advantageous, the subject matter has been considerably enlarged and numerous problems added. In conclusion I wish to express my thanks to the University of Toronto for making the publication of this book possible, and to Dr. Cecilia Krieger

for translating it from the Polish manuscript. WaAcLAWSIERPINSKI Warsaw,

October

1948

TRANSLATOR’S

PREFACE

YA J IEN a new edition of Introduction to General Topology was being considered, Professor Sierpifiski informed me that he had prepared a new manuscript on ‘General Topology” differing from the ‘‘Introduction’”’ in both content and treatment. He expressed the hope that the University of Toronto Press would publish a translation of the new manuscript in preference to a revised edition of the ‘‘Introduction.” The appendix appearing at the end of the Introduction is reprinted here with very slight changes. The numerous footnotes have, for economy in printing, been placed at the end of the book. For the same reason, the usual notation for analytic sets was changed. It is hoped that this change will not place any serious difficulties in the way of the reader. I wish to take this opportunity to express my deep gratitude to all those who with their discussion and criticism contributed to the enjoyment of a task which might easily have proved tedious. My special thanks are due to Mr. L. W. Crompton and Mr. W. T. Sharp who read part of the manuscript and to Dr. R. G. Stanton who read all of it and offered valuable suggestions. Cy CECILIARICRIEGER Toronto,

February

1952

Vii

.

—

uf

~

oe mw

Sa

CONTENTS

I. FRECHET

(V)SPACES

SECTION

PAGE

1.

Fréchet

(V)spaces

2.

Limit

3.

Topological

4.

Closed

5.

The

6.

Open

7.

Sets

dense-in-themselves.

8.

Sets

closed

9.

Separated

elements

5.

and

derived

equivalence

sets

3

of (V)spaces

4

sets

6

closure

of a set

sets.

The

7

interior

of a set The

in a given sets.

and

10.

Images

11.

Continuity.

Continuous

12.

Conditions

for continuity

13.

A continuous

14.

Homeomorphic

15.

Topological

16.

Limit

18.

Topological

sets of sets.

Biuniform

functions

images

22 24

in a set

26

of a connected

set

27 28

r

of order

m.

Elements

31 of condensation.

sets

33

theorem

34

limits

of a sequence

of sets

TOPOLOGICAL

19.

Topological

spaces

20.

Properties

of derived

sets

21.

Properties

of families

of closed

22.

Properties

of closure

23.

Examples

of topological

24.

Properties

of relatively

25.

Homeomorphism

26.

The III.

13 16

properties

II.

border

sets

sets

m-compact Cantor’s

of a set. Scattered

15

images

image

elements

17.

nucleus

set

Connected

inverse

11

36

SPACES 38 40 sets

43 45

spaces

47

closed

sets

in topological

of a set.

50

spaces

Nowhere-dense

TOPOLOGICAL

SPACES

50

sets

WITH

52 A

COUNTABLE

BASIS 27.

Topological

28.

Hereditary bases

spaces

with

separability

countable

bases

58

of topological spaces with countable 60 ix

CONTENTS

29.

power

of an aggregate

30.

The countability

OL;

The

Pa 33. 34.

IV.

The

of open

of scattered

Cantor-Bendixson

61

sets

62

sets

63

theorem

65

The Lindeléf and Borel-Lebesgue theorems Transfinite descending sequences of closed sets Bicompact

HAUSDORFF

66 69

sets TOPOLOGICAL

AXIOM

SPACES

OF

SATISFYING

THE

COUNTABILITY

of:

Hausdorff topological spaces. The limit of a sequence. Fréchet’s (Z)class Properties of limit elements Properties of functions continuous in a given set

38.

The

39.

given set. Topological types Continuous images of compact closed

40.

The

inverse

of a function

41.

The

power

of an aggregate

SEP

36.

power

V. 42.

Condition

43.

The

powers

of the

aggregate

NORMAL

FIRST

of functions sets.

continuous

of open

continuous

TOPOLOGICAL

76 78

in a

Continua

in a compact

(closed)

72

closed

set

sets SPACES

of normality of a perfect

compact

set and

a closed

compact

set 44.

Urysohn’s lemma

45.

The

power

of a connected

VI. 46. 47. 48. 49, 50. oie 52. Da: 54, sky 56. Die 58. 59. 60.

set

METRIC

SPACES

Metric spaces Congruence of sets. Equivalence by division Open spheres Continuity of the distance function Separable metric spaces Properties of compact sets The diameter of a set and its properties Properties equivalent to separability Properties equivalent to closedness and compactness The derived set of a compact set Condition for connectedness. ¢-chains Hilbert space and its properties Urysohn’s theorem. Dimensional types Fréchet’s space E,, and its properties The 0-dimensional Baire space. The Cantor set

105 106 107 109 110 115 117 119 120 122 128 133 142

CONTENTS

Closed

and

compact

sets

xi

as continuous

images

of the

Cantor 146

set Biuniform and continuous images of sets Uniform continuity Uniform convergence of a sequence of functions The (C) space of all functions continuous in the interval [0, 1]

158

The

space

163

. Sets F, and G; The straight line as the sum of Ni ascending sets G; Hausdorff’s sets P* and Q+ Sets which are locally P* and Q* Sets locally of the first category

166

.

space

of all bounded

Oscillation

closed

sets

of a metric

151 153 155

of a function

167 174 178 181 183

VU. COMPLETE SPACES Complete spaces The complete space containing a given metric space Absolutely closed spaces. Complete topological spaces The category of a complete space Continuity extended to a set G;

198

Lavrentieff’s

200

theorem

Conclusions

from

¥

Lavrentieff’s

theorem

A new

condition

Generalized .

The

207 208 210 213 214 216

.

Souslin’s Biuniform

sets. Topological

invariance 219

for a set to be analytic

of an analytic

220 Lao 224

set

228

theorem and continuous

images

of the set of irrational 232

numbers

The

property of a metric

.

The

theorem

197

205

sieves

power

195

204

for a set to be analytic

Continuous images of analytic of their complements .

192

202

Topological invariance of sets P* and Q* Borel sets: their topological invariance . Analytic sets; defining systems . The operation A. Lusin’s sieve Fundamental properties of the operation A Every Borel set an analytic set Regular defining system Condition

186

of compact

closed

or compact

open

subsets 235

space of Mazurkiewicz

about

linear

sets

G;

236

CONTENTS

xii 96. 97. 98. 99. 100. 101. 102. 103. 104.

Biuniform and continuous images of Borel sets The analytic set as a sum or an intersection of &: Borel sets Projection of closed sets Analytic sets as projections of sets G3 Projective sets Universal sets Universal sets P, and C, The existence of projective sets of any given class The universal set F,5

240 244 248 249 250 251 Zaz 253 254

APPENDIX

256

NOTES

279

INDEX

289

=o

#

CHAPTER I FRECHET 1. Fréchet set

(V)spaces.

K of elements

class

of subsets

(V) SPACES

A Fréchet

in which of K called

with

each

(V)space, element

neighbourhoods

or briefly a there

a (V)space,

is associated

is a

a certain

of a.

Thus the set of points in the plane is a (V)space if a neighbourhood of a point p is taken to be, e.g., the interior of an arbitrary circle with centre at p. Clearly, a neighbourhood in this case can be defined in many ways as, for instance, the interior and boundary of any square with centre at p. It would also be consistent with the definition to assume that each point p of the plane possesses only one neighbourhood, e.g., the set consisting of p itself. The set of all real functions of a real variable is a (V)space, if a neighbourhood of f(x) is defined to be the set of all functions g(x) which, for a given positive c and for all values of x, satisfy the inequality

If(x) — g@)|

w (§ 35).

CoNnTINUOUS

A function pb © Eif,

f(p)

is said

for every

to have

infinite

limit

sequence

IMAGES

25

continuity

/1, p2,...

in a set E at the element for which

lim p, = p, we have nN—-co

lim f(fn) no according converse, about the

= f(p).

function

continuous

at an element

to our definition has also limit continuity at that element. The however, need not be true unless certain assumptions be made space K (see § 37).

It follows constant

It is easily seen that a function

from

in £

the

definition

is continuous

is continuous

of continuity at

every

that

element

in a set £ consisting

a function

of #.

of a single

In

f(p)

which

particular,

is

every

element.

The set of all real numbers is a (V)space if by a neighbourhood of the number x) we mean the interior of any interval containing x». Here, as is easily seen, our definition of continuity is identical with that used by Cauchy in defining a continuous (real) function of a real variable. If a function f is continuous at every element of a set E, it is said to be continuous in £, and the set T = f(£) is called a continuous image of E obtained by means of the function f. Obviously, if po € E: C E and if f is continuous in £ at fo, then f is also continuous in EF; at fo.

THEOREM 21. Jf the function f 1s continuous in E at po and the function g is continuous in T = f(E) at go = f (fo), then the function 6(p) = g(f(p)) ts continuous in E at po. 7

Proof. Let W be a neighbourhood of ¢(o) (in the space to which this element belongs); since ¢(fo) = g(f(o)) and gis continuous in T at f(po), there exists a neighbourhood V (corresponding to W) of f(fo) (in the space to which f(p0) belongs) such that

(45)

g(q)

€W

fongre

Veels

Since f(p) is continuous in E at fo there exists a neighbourhood U (corresponding to V) of po such that (44) is true. From (45) and (44) we obtain at once

(46)

g(f(o))

€W

forp €U.E.

Thus for every neighbourhood W of $(o) there exists a neighbourhood U of fo for which (46) holds; since ¢(p) = g(f (2), this proves that the function ¢ is continuous in E at po. In particular, if f is continuous in the whole set £ and g is continuous in the whole set T = f(£), then the function ¢ is continuous in the whole set E. In other words, a continuous image of a continuous image of a given set is a continuous image of that set.

26

FRECHET (V)SPACES 12.

Conditions

for continuity

in a set.

THEOREM 22. A function f defined in a set E is continuous in E af and only if the image of any element of E which is a limit element of any subset of E 1s an element or limit element of the image of that subset.

In other words, for f to be continuous it is necessary and sufficient that (47) f(E1.E) Cf(E:) + GE), ErGE. Proof. Let f be a function defined and continuous in E, £; a subset of £, and pp an element of Z’;. E. Assume that f(po) ¢f(£1) and let V denote a neighbourhood of f(fo). Since f is continuous in E at po there exists a neighbourhood U of o for which (44) holds. From fo € £/1.£ C £’,, it follows that U.H, #0; there exists, therefore, at least one element p€ U.E, C U.E such that f(p) € f(FZ:) and, from (44), f(p) € V; since f (bo) ¢f(E1), we have f(p) # f(po). Thus every neighbourhood of f(o) contains an element of f(£;) different from f(o); consequently f(f0) € (f(A1))’. This proves the necessity of the condition. Next assume that f is a function defined in E but not continuous at fo € E. Hence there exists a neighbourhood V of f(0) such that every neighbourhood U of fo contains at least one element p € U.E for which f(p) ¢ V. Let Fy denote the set of all elements p of E for which f(p) ¢ V; then po ¢ Ay since f(po) € V, and therefore f(fo) ¢f(£1); also f(fo) ¢ (f(E:1))’ since F(Ex) (V —f(f0)) = 0. Consequently (47) is not true. The condition of the theorem is therefore sufficient and the theorem is proved. Let f be a (1,1) function, continuous in a set E. Let FE; be a subset of E, po an element of E . £’;, and V a neighbourhood of f(fo). Since f is continuous in £, there exists a neighbourhood U of fo for which (44) holds. But bo€ E. FE’;C £1; hence there exists an element » € U.£, such that pb~ po; then f(p) € f(Z1) and, since f is (1,1), f(p) ¥ f(p0). Thus every neighbourhood V of f(/0) contains an element of f(£:) different from f(po); consequently f(po) € (f(41))’. We have

therefore

(48)

JE

EYVCUED

forall Ey C

On the other hand, if a function f defined in a set E satisfies (48), it also satisfies (47) and so, by Theorem 22, it is continuous in E. We thus obtain:

THEOREM 23. In order that a function f, defined and (1,1) in a set E, be continuous in E, tt is necessary and sufficient that

FEED). From

Theorem

CoroLiary.

22 we obtain

the

Cay:

for all Ey C E.

following

If a function f is continuous in a set E and T is a set closed in

ConTINUOUS

IMAGES OF CONNECTED

SETS

27

T = f(£), then the set Ey = E[f(p) € Ti], that ts, the set of all elements p of E for which f(p) € Ti, ts closed in E.

Proof. Let p € E. E’;; since f is continuous, by Theorem 22, f(p0) € f(F:) + (f(Ei))’. If f(po) € f(F1), then po € E:; and if f(po) € (f(E1))’, since Po € E, f(bo) € Tif(A1))’ C T.T’1 C Ti, because T1is closed in T. Hence in this case too, po € E;. Consequently E . E’; C Fi, that is, EZ;is closed in E£. In particular,

if 7, C T, we can state

THEOREM 24. The inverse image of a set contained and closed in a continuous image T of a set E ts closed in E 13. A continuous

image

of a connected

set.

THEOREM 25. A continuous image of a connected set is connected. Proof. Let E denote a connected set, T = f(£) its continuous image, and assume that T is not connected. By Theorem 13, T = A; + Bi, where A; and B, are non-empty disjoint sets, each closed in 7. Put A= E,[f(p) € Ail, B=E,|f(p) € Bij; the sets A and Bare obviously non-empty, disjoint, and, by Theorem 24, each closed in £; moreover, EK = A + B and, since E£ is connected, this is impossible. Hence T is connected. Theorem under

25 may

continuous

be restated

as follows:

connectedness

of sets is an invariant

transformations.

The set of all real numbers is a (V)space if by aneighbourhood of a number x we mean any interval!! < a, b > where a and 0 are any numbers such that am; hence TJpossesses a limit element of order > m; consequently E is m-compact.

17. Cantor’s theorem. THEOREM 29, If (53)

DR eeOSDotZa

er

is an infinite descending sequence'® of non-empty closed sets (contained

in a

Limit

ELEMENTS

OF ORDER

(V)space) at least one of which is So-compact, 1s not empty.

m

35

then the intersection of these sets

Proof. Let E, be the first So-compact set in the sequence. Let a, denote an element of E, for 7 = 1, 2,..., and let P be the set of all a, for n > p. From (53), PC#,. If P is finite, one of the elements of the sequence Gy,Ap41, - . . 1Srepeated from a certain stage onwards; this element is obviously contained in F, for nm= 1, 2,... and, therefore, in the intersection II} E,,. On the other hand, if P is infinite, it possesses a limit element, say a, of order > No, since it is a subset of an No-compact set. Let V bea neighbourhood of a; it therefore contains an infinite number of elements of P and, since PCE, C Ey,” < p, V contains an infinite number of elements of E,, 2 < p. But only a finite number of elements of P may be outside the set E, for n > p; hence V contains an infinite number of elements of £, form = 1,2,... Therefore a € EL’, and, since E, is closed, a € E, form = 1, 2,...; hence ave This

proves

[i] 2,,. 1

the theorem.

The sequence of sets (53) in Theorem 29 cannot be replaced by a transfinite sequence (unless certain assumptions be made about the (V)space in which the sets are contained). For example, let K denote the set of all ordinal numbers of the first and second classes. K becomes a (V)space if the interior of any segment of ordinal numbers (i.e., all ordinal numbers &satisfying the relation a < —< 8) containing the number 7 + 1 is regarded as the neighbourhood of n, and if the neighbourhood of 1 is the set of all ordinal numbers < a, where a is any ordinal such that 1 So and that therefore every compact set 1s No-compact. Hence, for topological spaces, we may replace in Theorems 29, 30, and 31, the expression No-compact by compact. From

Theorem

33 (and

the

fact

that

the

null

set is a subset

of every

set)

we obtain

CorROLiary 3. Every finite set 1s closed or, in other words, every finite set is tdentical with its closure.

CoROLLARY4. Every connected set containing dense-in-ttself.

more than one element is

To prove this, let H denote a connected set containing more than one element and let p be an element of E. Put A = {p} and B = E — A; then A’ = 0, by Theorem 33. If we had A.B’ = 0, the sets A and B would be separated, contrary to assumption that E A.B’ #0; hence p € B’ and so certainly hence £ is dense-in-itself.

From Corollary 4 we conclude further taining more than one element 1s perfect.

is connected. p € EF’. This

Consequently gives EC E’;

that a connected and closed set con-

A set E; C E is said to be compact in the set if every infinite subset of E; possesses at least one limit element contained in £, in other words, if for every infinite subset E, C FE; we have E’;.E #0. In particular, a set E is said to be compact-in-ttself if every infinite subset of E possesses at least one limit

PROPERTIES

OF FAMILIES

OF CLOSED

SETS

43

element which is contained in Z. A set which is compact-in-itself is obviously compact. It need not, however, be closed as is illustrated by the following example. Let

K

denote

the

set

of all

ordinal

numbers

0) of elements of P we put

E = ¢(K.E)+P.E.

Hence, in particular, # = $(£) for all ECK. contains an infinity of elements of P, we put

If, on the other

hand,

E

Bie Ky: It can easily be shown that the closure as defined in K, satisfies the conditions (1), (II), and (III). K;, is therefore a topological space. It is evident from the definition of closure in K, that, in particular, P = IKE Hence K, is separable. The above example shows that every topological space K is a subset of a separable topological space obtained from K by adjoining a countable set. If K is a non-countable set in example 2 then P = K is impossible for any finite or countable subset P C K (for P = P and P has cardinal < No whereas K is non-countable). But, as shown above, K is a subset of a certain topological and separable space Ki; hence Ky, is not hereditarily separable (cf. example 3). There exist, therefore, separable topological spaces which are not hereditarily separable. 5. Let K denote any denoted by ¢(£). Let the closure # in the set If E is finite, we put

topological space in which the closure of a set £ is a be a given element not belonging to K. We define K, = K + {a} as follows: Z = E; if E is an infinite subset of Ki, we put

E = $¢(E — {a}) + {a}. As a consequence of the definition of closure in Ki, we have o(E)

=E.K

for alvEnAGKk

(which is consistent with the definition of closure in a subspace). We see that the closure in Ky, as defined above, satisfies conditions (1), (II), and (III); Ky, is therefore a topological space. We shall now show that Ky, is compact. Let E denote an infinite subset of Ki. Put E; = E — {a}; hence Zi is infinite and so, by the definition of closure in K1,a € E,. But £, = FE, + E; and a ¢ E,; hence a € E£’; C FE’; consequently E’ #0. The space Ki is therefore compact.

50

TOPOLOGICAL

SPACES

Thus every topological space K is a subset of a compact topological space obtained from K by adjoining a single element.

24. Properties of relatively closed sets. THEOREM39. A subset of ECK (K a topological space) which 1s closed in E is the intersection of E and a closed set of K and conversely. Proof. If Ei C E and &E;is closed in E, then E. ££’; C E; (§8); so E, = (£,+ E')E = B,.E. Hence £; is the intersection of E and the closed set FE. On the other hand, if E; = F.£, where F is closed, then E; C F and so E'; C F; therefore E’;.EC F.E = E,. Hence £; is closed in £ and this proves the theorem. In the general (V)space only the second part of Theorem 39 holds.

25. Homeomorphism in topological spaces. By Theorem 27, two (V)spaces K and K, are homeomorphic if and only if there exists a (1,1) mapping f of K on K,; such that (11) (2) =f) for allE CK.

We now show that if K and K, are topological spaces (11) is equivalent to (12) Proof.

f(E)

= f(£)

forallE

CK.

If we assume (11), then (1) gives

f(E)

=f(E)

+ ((2))

=f)

+fE)

=fE+

£)

=f)

which proves (12). Assume that (12) is true. Take any 0 € f(Z’). We then have db= f(a) for some element a € EZ’. Put £; = E — {a}. Since E’; = E’, we see that a € E';C Ey. Hence bd= f(a) € f(F) and so, by (12), b € f(E:)

= f(E:)

+ (f(Ey)y’.

Since f is (1,1) in K, anda ¢ FE, we see that 6 = f(a) ¢ f(E:). Thus we have b € (f(A1))’ and, since Ei C FE,b € (f(Z))’. Therefore relation (12) implies that f(E’) C (f(2))’. Suppose next that b € (f(£))’. Since b € K, = f(K), there exists some element a € K such that 6 = f(a). Put A, = E — {a}. Since f is (1,1) in K, f(E:) = f(E) — {b} and we see that 6 ¢ f(E:). Thus, since (f(£,))’ = (f(E))’, we have b € (f(E:)) C f(A). By (12), 6 € f(#:) = f(E:+ £41) = f(Ei) + f(E'). But b ¢ f(Ey); so we know that b € f(Z1) Cf(Z’). Therefore (12) implies that (f(Z))’ C f(E’). Hence (f(£))’ = f(Z’) and relations (11) and (12) have been shown to be equivalent. We have therefore proved

HOMEOMORPHISM

IN TOPOLOGICAL

SPACES

Bil

THEOREM 40. Two topological spaces K and K, are homeomorphic if and only wfthere exists a (1,1) mapping of K on K, such that the closure of the image of any set contained in K 1s the image of the closure of this set. In other words, the closures of corresponding sets under a (1,1) transformation are corresponding Sets. It follows from Theorem 40 that if two topological spaces are homeomorphic it is possible to establish a (1,1) correspondence between their elements such that the image of a closed subset of either space is always a closed subset of the other. This is true in every (V)space—see remark to Theorem 27, § 14. We now prove the converse, which need not hold in the general (V)space. Suppose that f establishes a (1,1) transformation of the topological space K on the topological space K; in such a manner that if E is a closed subset of K then f(£) is a closed subset of K; and if T is a closed subset of K, then f-'(T) is a closed subset of K. Hence if Ey be any subset of K, then each closed set E containing /» is mapped into a closed subset of K, containing f(£o) and the function f~! transforms every closed set containing f(£o) into a closed set containing Ey. Since f is (1,1) in K, it transforms the intersection of all closed sets (contained in K) which contain Ep into the intersection of all closed sets (contained in K,) which contain f(£o) i.e., it transforms the set

E, into the set f (Eo). We have therefore f(E) for all ECK; hence, by Theorem thus obtain the

= f(£) 40, K and K, are homeomorphic.

We

CoROLLARY. Two topological spaces K and K, are homeomorphic tf and only af there exists a (1,1) correspondence between their elements such that closed sets of K correspond to closed sets of Ky and vice versa. It should be noted, however, that in order that two topological spaces K and K,; be homeomorphic it is not sufficient that there exist between their elements a (1,1) correspondence under which closed sets of K correspond to closed sets of K;. The theorem fails if closed sets of K, map on sets of K which

are not closed.

For and

example,

K;, is the

number

2, then

if

K and

spaces.

Also,

f(1)

= 2 transforms K is connected

into

the

a set not

of all numbers

the sets

topological

is because

the set of all real

set consisting

are since

K denotes

inverse closed

every and

closed

the

x, where

as subsets function

subset

K; is not,

function in K.

K,,

the

x, where0

< x
2 such that p, € EL and so the set E = {p,} + (E — {p,}) is the sumof two separated sets. It is also easily

verified

that

;

and f2 are not separated

in K.

CHAPTER

TOPOLOGICAL 27.

Topological

to possess

SPACES

spaces

a countable

basis

(1)

with

III

WITH

countable

if there

A COUNTABLE bases.

exists

A topological

an infinite

BASIS space

K is said

—

sequence

Ui, Us, Us...

of open sets such that every open set contained in K is the sum of a certain aggregate of sets belonging to the sequence (1). The sets of the sequence (1) will be called ratzonal sets. We shall deduce in this chapter a number of theorems for topological spaces — with a countable basis. It is evident from the Corollary to Theorem 40 that the existence of a countable basis in a topological space is a topological property.

LewMa 1. Ifa isa given element belonging to K, and U an open set containing a, then there exists a rational set containing a and contained in U. Proof. sets, i.e

The set U, being open, is the sum of a certain aggregate

M of rational .

U= UieM

hence

if a € U,a

€ U; C

U for some

k.

This

proves

the lemma.

LemMA 2. Corresponding to every element a belonging to K there exists an infinite sequence V1, V2,... of rational sets such that their intersection consists of the element a only, and every open set V containing a contains all but a finite number of the sets Vi, Vo,.... Proof.

Let a be a given element io.

of the space K and let Dy, see

be the sets of the sequence (1) which contain a in the same order as they appear in (1). (It may always be assumed that this sequence is infinite by repeating the last term infinitely many times if necessary.) Let k be a given’ natural number. The intersection tows U,, see De. is, by Theorem 36, an open set and it contains exists a rational set V; such that 58

a; hence, by Lemma

1, there

TOPOLOGICAL

SPACES WITH COUNTABLE

Cee Coe

BASES

59

Us

(It may be assumed that V; is the first set of the sequence (1) which satisfies the above condition.) We shall show that the sequence {V;},k = 1,2..., satisfies the conditions of Lemma 2. For, on the one hand, a € V; for k=1,2,...; andif V is an open set containing a then, by Lemma 1, there exists a rational set, say U,,, containing a and contained in V. Hence for k > q we have, from the definition of the sets V;,

Vee

Ue GW.

and so V;, C V fork > gq. On the other hand, if } is an element of K different from a, there exists, by condition y and the fact that in topological spaces neighbourhoods are open sets, an open set V containing a but not containing b. Hence the sets V;, which are contained in V for all sufficiently large k, do not contain b; consequently the intersection V;. V2. V3... contains no element different from a. Lemma 2 is therefore proved. Moreover, it can be proved that the sequence Vi, Vo,...of Lemma 2 may be assumed to be descending, that is, SS Ce 2 ee ae In connection with Lemma 2 we remark further that the following condition W is called the first axiom of countability (Hausdorff): Condition

W.

Corresponding

to every element

a of a space

K there exists

an

infinite sequence V1, Vo,... of open sets containing a and such that, for every open set V containing a, all but a finite number of the sets Vi, Vo,... are contained in V. It follows from Lemma 2 that every topological space with a countable basis satisfies the first axiom of countability. There exist, however, topological spaces which satisfy the first axiom of countability but do not possess a countable basis. For example, the non-countable space K defined in example 2 of §23is such aspace. The condition that a countable basis exist is called by Hausdorff the second axiom of countability. From

Lemma

THEOREM 41. Proof. intersection

It

2 we obtain

A topological

follows

from

of all those

space

Lemma sets

with a countable

2 that

every

of the sequence

N@)=Elae (i.e., N(a) is the set of all natural

element

(1) in which

Vn]

numbers

has cardinal

of the

space

it is contained.

< €.

K is the Put

a€kK

x for which a € V,); obviously

{a}= 1m)Vi. neN (a)

basis

60

TopOLOGICAL

SPACES

WITH

A COUNTABLE

BaAsIs

Every element of K is therefore completely determined by the set V(@); that is, by a certain set of natural numbers. Since the aggregate of all sets of natural numbers has cardinal ¢, it follows that K 1 has only one neighbourhood consisting of the number itself; the number 1 has for its neighbourhoods all sets V containing 1 and such that

a

N(n, V) “

naa

where

NV(n,V)

denotes

the

number

1,

n of those

numbers

in the

set

V which

POWER

OF AGGREGATES

OF OPEN

SETS

61

are 1 has only one neighbourhood consisting of the number itself. The number 1 has its neighbourhoods defined as follows: a set Visa neighbourhood of the number 1 if it contains 1 and if there exists an infinite sequence 1, l2,... of ‘natural numbers such that V is the set of all natural numbers of the form 2*(21 — 1) where 2 and / are natural numbers and / > i.

It is easily verified that S is a topological space. Let now V1, Vo,... denote any infinite sequence of open sets containing 1 and contained in S and let k be a given natural number. The number 1 is obviously a limit element of the set of numbers {2*(2/ — 1)}, / = 1, 2,... (since every neighbourhood of 1 contains infinitely many numbers of this set). There exists, therefore, a natural number /;, such that the number 2*(2/, — 1) € V;. Let V denote the set of all natural numbers of the form 2*(2/ — 1) where k = 1, 2,... and 1> 1, Hence V is a neighbourhood of 1 and consequently an open set. But my —1) ¢ V for k= 1, 2,... and so V, — V0 tor k=1,.2,.... Thus no infinite sequence V1, V2,... of open sets satisfies condition W of § 27 for the element 1. The space S does not satisfy the first axiom of countability; consequently it does not possess a countable basis. However, every countable (V)space which satisfies the first axiom of countability possesses a countable basis. Proof. Let K denote a given countable (V)space which satisfies the first axiom of countability. Hence, corresponding to every element a € K there exists an infinite sequence V,(a), m= 1, 2,..., of open sets which satisfies condition W of §27. Since the cardinal number of K is No, the family B of the sets V,(a), a € K,n = 1, 2,..., is countable. Moreover, the family B forms K,

a countable

then

U is the

at once

from

the

that

V,(a)

some

set of the

29.

basis

The

C

U;

of the

space

sum

of all sets

of

fact

that

B which

For

B which

a € U then

consequently

family

power

if

K.

every

Let

K denote

of open

let U be an open set for which U, C U.

a topological

exists

element

THEOREM 43. The aggregate of all open with a countable basis has cardinal < c. Proof.

are contained

there

is contained

of an aggregate

if U is an open

space

set contained

in U.

a natural

of the

set

This

follows

number

7 such

U is contained

in

in U. sets.

(closed) sets in a topological

with

the countable

basis

C K. Denote by N(U) the set of all natural It is evident from the definition of the basis

C=

in

space

(1) and

numbers (1) that

”

Uy. neN(U)

Hence every open set U C K is uniquely determined by a set of natural numbers i.e., by N(U); consequently the aggregate of all open sets of K has

62

TOPOLOGICAL

cardinal

) (E;—Ean), 0 No, we have altogether 2*™ different topological types. 39. Continuous

images

of compact

closed

sets.

Continua.

THEOREM57. A continuous image of a closed and compact set contained in a Hausdorff topological space satisfying the first axiom of countability and itself contained in such a space, ts closed and compact.* Proof. Let E be a given closed and compact set which is contained in an (HT)space K satisfying the first axiom of countability. Let f(p) be a function defined and continuous in E whose values belong to an (HT)space K;, satisfying the first axiom of countability (where we may have A; = &). Let 7; denote any infinite subset of the set T = f(£). There exists, therefore, an infinite sequence {g,} of different elements of 7}. Since

@EeEMCT=f*),

n= 1,2,...

there exists, for every natural number m, an element p, € E such that Qn = f (bn) and all the terms of the sequence {/,} are different (since the terms of the sequence {g,} are all different). Denote by 2p the set of all the terms of the sequence {f,}. Then £ is an infinite subset of EZand so has a non-empty derived set, since E is compact. Let a be an element of E’o. Since Ey) C E and E£ is closed, we have a € E and so f(a) € f(Z) = T. Let V denote any neighbourhood of f(a) in Ky. Since f is continuous in F, there exists a neighbourhood U of a such that f(p) € Vforp € U.E. Sincea € E’o, there exists, by Corollary 2 to Theorem 33 (§ 20), two different elements », and dn in the set Eo. U. Hence gn = f(bm) € V and gn = f(pa) € V. Since m #* nwe have dm ~ g, and so at least one of the elements, say @m,is different from f(a). Thus in every neighbourhood V of f(a) there is at least one element of 7, different from f(a); consequently f(a) € T’;, that is 7’; ¥ 0. Since 7) is any infinite subset of 7, it follows that T is compact.® Now let } denote any element of A; such that 6 € T’. (and the remark to it), there exists an infinite sequence

By Theorem 53 {g,} of different

ContTINuous

elements

of the

set

T such

(11) Since number

IMAGES oF

Compact

CLOSED

SETS

83

that

lim gy = 6. g¢, € T =f(£) z an element

for » = 1, 2, ..., p, such that

there

exists

for

every

natural

Pn ©Eand f(p,) = Ins where the terms of the sequence {,} are all different. Denote by EZ»the set of all terms of the sequence {,}. is an infinite set contained in the compact set E; hence E’) #0. There exists, therefore, an element a € EF’. Hence, by Theorem 53, there exists an infinite sequence of elements of Eo, all different from a, whose limit is a. This sequence differs at most in the order of its terms from a subsequence of the sequence {p,} and, since the limit of a sequence is independent of the order of its terms (§ 35), we may conclude that there exists an infinite subsequence {,,} of the sequence {p,} such that

(12)

lim py, = a. k-yc0

Since a € E’y and E£ is closed, we know that a € E; E, it follows from (12) and Theorem 54 that

since f is continuous

in

lim f(Pa.) = f(a) 1e., lim gu, = f(a). kn kw As a consequence of properties 2 and 3 of a limit (§ 35) we have b = f(a) and so (sincea € E)b € f(E) = T. Hence bd€ T”’implies b € T, i.e., T is closed. Theorem 57 is therefore proved. We note that in an (HT)space with a countable basis (e.g., in the set of all real numbers) a continuous or even a homeomorphic image of a closed set need not be closed. Thus, for example, a finite open interval of real numbers (not a closed set) is a homeomorphic image of the set of all real numbers (a closed set). Similarly, a continuous or even a homeomorphic image of a compact set need not be compact; e.g., the set of all real numbers (not compact) is a homeomorphic image of a finite open interval of real numbers (a compact set). Moreover, a homeomorphic image of a closed or a compact set contained in a topological space may be neither closed nor compact. For example, the set of all positive real numbers is neither closed nor compact yet it is a homeomorphic image of the closed set of all real numbers as well as a homeomorphic image of the compact set of all real numbers in the open interval (0, 1).

Furthermore, in an (H7)space which does not satisfy the first axiom of countability a homeomorphic image of a closed and compact set need not be closed.

84

HauspDoRFF

TOPOLOGICAL

SPACES

For example, let K denote the set of all ordinal numbers 0. A neighbourhood of the point p = (x, y), y > 0, is the interior of any circle 90

CONDITION

OF NORMALITY

91

with centre at p and radius < y. A neighbourhood of p = (x, 0) is a set consisting of p and the interior of any circle tangent to the x-axis at p and having its centre above the x-axis. It can be shown (the proof is rather difficult) that the set P of all the points of the x-axis with rational abscissae and the set Q of points on the x-axis with irrational abscissae are closed disjoint sets of K but that they cannot be covered by two disjoint open sets U and V such that UDPand VD Q. On the other hand, it is easy to give an example of a Hausdorff topological space which is not normal. Such a space is, for example, the set of all real numbers x of the interval 0 < x < 1, where a neighbourhood of the number x is any set consisting of x and all the rational numbers of the above interval which differ from x by less than 1/n for m any natural number. Furthermore, it can be shown that a regular topological space with a countable basis is normal.? Problems

1. Let T denote the set of all real numbers. Define the closure X of a set X CT to be the set obtained by adjoining to X all its elements of condensation (in the ordinary sense). Prove that T is a Hausdorff topological space which does not satisfy the first axiom of countability nor the condition of regularity. 2. Prove that if a topological space T is normal and P and Q are two closed sets contained respectively in two disjoint open sets U and V (contained in T), then there exist open sets U, and V; such that PC Wi, Ui C U, and OCV,ViCV. 3. Prove that if the sets P and Q contained countable basis are closed, compact, and disjoint, U and V such that PC U,QC V,and U.V=0.

in an (HT)space with a then there exist open sets.

Proof. Since T is an (HT)space and so satisfies condition 1, there exist for every pair of different elements » and g open sets U(p, g) and V(p, qg)such that p € U(p, g), g € Vip, gq),and U(f,q) . V(p,q) = 0. Let p be a given element of P and R(p) the family of all open sets V(p, q) for g € Q. The closed and compact set Q is therefore covered by the sets of the family R(p) and, since T possesses a countable basis, it follows from the Borel-Lebesgue Theorem (§ 32) that there exists a finite set q1, G2,..., Gm,of elements of Q such that 0 C V(b, g1) + V(b, go) +... + V(b, Gm) = V(p). The set V(p) is obviously open and, since U(p, gi) . V(p, qi) = 0 forz = 1, 2,..., m, the open set U(p) = U(p, qi)... U(~, Gm) contains p and has no elements in common with the set V(p). The set P is clearly contained in the sum of all sets U(p), where » € P; consequently there exists, as in the case of the set Q, a finite sequence 1, po,..., py of elements of the set P such

92

NorMAL

TOPOLOGICAL

SPACES

that P C U(p1) + U(p2) + 2.. + U(fn) = U. Since U(p) . V(b) = 0 for pb€ P, we have, for V= V(p1). V(p2)... V(pr), U. V=0. The sets U and V are open and satisfy the required conditions. 4. Show that a compact Hausdorff topological space is normal. The proof follows from the result of problem 3 and the fact that compactness of a set is a hereditary property. 43. The powers

of a perfect

THEOREM 60. Every normal

compact

non-empty,

space has cardinal

set and a closed

perfect,

and

compact

compact

set.

set contained

in

a

> C.

We shall first prove the following Lemma. Corresponding to every non-empty perfect set P contained space, there exist two non-empty perfect sets Py and P; such that

(1)

Peer,

Pie

Pans

in a normal

fi

Proof. Let P denote a given non-empty perfect set contained in a normal space K, po and p; two different elements of P (such elements exist since a non-empty perfect set cannot consist of only one element). Since K is normal, there exist open sets Up and U, such that bo € Uo, p1 € Ui, and

Uy. U, = 0.

But the condition of normality implies the condition of regularity (§ 42); hence there exist open sets Vo and V; such that po € Vo, £1 € Vi, Vo C Us, Vi C U;. Furthermore, since Up. U; = 0, we have Vo. Vi = 0. Since bo € Vo. P and P is perfect and hence dense-in-itself, it follows that Vo. P is dense-in-itself (see the proof of Theorem 47). The set fy

= Vote

0

is dense-in-itself by Theorem 6; hence it is perfect. and Py C Vp (since Vo. PCPand Vo. PC Vo).

Moreover, Py) CP = P

Similarly, we conclude the existence of a non-empty and P:C Vi. Since Vo. Vi = 0, we have Py.P:;=0. lemma.

perfect set P; C P This proves the

Since Up may be considered to be a subset of an open set containing po, we have actually proved that every open set containing an element of a perfect set P contains a non-empty perfect proper subset of P. To prove Theorem 60 let P denote a non-empty perfect and compact set contained in a normal space. By the above lemma there exist perfect sets Py and P; which satisfy condition (1).

POWER

Let &be a given natural

OF PERFECT

COMPACT

SETS

93

number and suppose all the non-empty

perfect sets

Paves. ick already defined, where aia2...a, is an arrangement of & numbers each of which is either 0 or 1. By our lemma, there exist non-empty perfect sets Poies.0s0

ANd Po,a,...431

which satisfy the following conditions:

(2)

Praaeke ce darentcas,Teatreet GiSBaekeso

(3)

ates

em er0 . Periea

Se 0.

The sets Pa,a,---a,, Where aja2...a, is any finite arrangement numbers 0 and 1, are thus defined by induction. Now let (4) be

of the

Qi, Q2, 43,-.. an

infinite

sequence

formed

with

the

numbers

0 and

1.

Consider

the

intersection

(5) The sets

fia itebeta sassPease; eciauteens Tia:

03

(k = 1,2,...)

form, by (2), an infinite descending sequence of non-empty sets. Furthermore, as subsets of the compact set P, they are compact and, since they are perfect, they are also closed. Consequently, by Theorem 30 (Cantor), the intersection (5) is not empty. Let (a1, a2,...) denote an element of this intersection. Thus to every infinite sequence (4) formed with the numbers 0 and 1 corresponds an element p(a1, a2,...) of theset P. There are c such sequences. ‘Moreover, to different sequences (4) correspond different elements of P. For let B1, B2,..+ denote from denote

an the

infinite sequence

the

smallest

sequence (4);

then

of them.

(6)

formed

with

there

the

numbers

exist

indices

a,=

B;

0 and

m such

that

1 and a, #

different

Bn.

Let

m

Hence

fordinrl,

22. ae3 70 —1;

but am ~ Bm; assume (7) dm = 0, Bm = 1. It follows from the definition of the elements (aj, a2,...) that (8) p (a1, Q2,+- €. This completes the proof. Let p be any element of a perfect and compact set P and let U be an open set containing p. As shown in the preceding lemma, the intersection P . U contains a non-empty perfect subset. Furthermore, this subset is compact; hence, by Theorem 60, it has cardinal > c. Thus:

If p is an element of a perfect and compact set P contained in a normal space, then every open set containing p contains a subset of P of cardinal > c.

This gives immediately CoroLiary 1. Every element of a perfect, compact set contained in a normal space 1s an element of condensation of the set. Theorems

41 and

60 lead

to

CoROLLARY 2. In a normal perfect, compact set has cardinal From

Corollary

2 and

Theorem

space c.

with

a countable

basis

every non-empty

47 we obtain

CoROLLARY 3. In a normal space with a countable closed and compact set has the power of the continuum,

basis every non-countable 1.e., cardinal c.

We note that the condition of compactness cannot be omitted in Theorem 60 or in Corollaries 1, 2, and 3. In fact, the set of all rational numbers (with the usual definition of neighbourhoods) is clearly a normal and perfect but countable space.

Nor can the condition of normality be omitted from Theorem 60, as can be proved by the example of a countable topological space T in which the closure of every infinite set is the whole space T. The condition of normality may, however, be replaced by the condition of regularity, as is easily seen from the proof of Theorem 60. Furthermore, in Corollaries 2 and 3 the condition that the space be normal may be replaced by the condition that. the space be an (HT)space. This follows from the result of problem 4 and the fact that a subset of an (HT)space is an (HT)space.

Urysoun’s

LEMMA

95

Finally, the inequality in Theorem 60 cannot be replaced by the sign of equality. For, given any cardinalm > ¢, it is possible to give an example of a normal continuum (which is by definition a perfect, compact, and nonempty set) of cardinal m. Thus, for instance, let ¢ denote any ordinal number >0 and m its cardinal. Denote by M the set of all pairs (é, f) where € is an ordinal number satisfying the inequality 0 < —< ¢ and ¢ is a real number such that 0 < ¢ < 1, and the pair (¢,0). The set M will be ordered if we assume that (&1,t1) < (2, fo) if & < £ and, where & = &, if 4) c, it has cardinal m. Hence there exist normal spaces which are continua of cardinal > c. 44. Urysohn’s

lemma.

Urysoun’s Lemma. Jf P and Q are two disjoint closed sets contained in a normal space K, then there exists a real function f(p) defined and continuous in K such that 0 @ — 1)/2" =1—1/2"* > t—e On the other hand, we always have f(p) < 1; hence 0 < f(o) The function is therefore continuous in K at po.

—f(p)

< e.

Case 3. 0 c.

set consisting

of more than one element

if U 1s an open set such that

and

S.U # 0, then

It is noteworthy that (under the hypothesis of Theorem 61) the set S.U need not be connected and it need not even contain any connected subset consisting of more than one element.* Theorems 61 and 41 give CoROLLARY 1. Every connected set consisting of more than one element and contained in a normal space with a countable basis has the power of the continuum.

There exist, however, as stated (without proof) by Urysohn,‘ countable connected topological spaces with countable bases. It is easy to see that the space K of example 3, § 23, is such a space for m = No. Since a continuum is a closed, connected, and compact set containing more than one element (§ 39), we obtain, in conjunction with Theorems 61 and 41, Coro.iary 2. If a set C 1s a continuum contained in a normal space with a countable basis, then the intersection of every open set with C, if non-empty, has cardinal c.

This property

justifies the term ‘“‘continuum.”

CHAPTER

METRIC

VI

SPACES

46. Metric spaces. A set M is a metric space (Hausdorff) if with every pair of elements a and b of M there is associated a real non-negative number! p(a, b) called the distance between the elements a and 0 and subject to the following three conditions (so-called distance axioms):

1. p(b, a) = p(a, 6) (symmetry law); 2. p(a, 6) = O if and only if a = d (identity law); 3. p(a, c) < p(a, 6) + p(b, c) for every three elements a, b, c of M (the triangle law).? A subset of a metric space is evidently a metric space. Elements of a metric space are frequently called points. Menger? calls a space M semi-metric if with every pair of elements of there is associated a non-negative number subject to conditions 1 and 2 (but not necessarily 3). Semi-metric spaces have also been investigated by Chittenden.* There exist semi-metric spaces which are not topological and there are topological spaces whose homeomorphic images are not semi-metric spaces.°® Even more general spaces have been investigated in which the distance function satisfies neither the triangle law nor the symmetry law.® Alexandroff and Hopf? call a space in which there is defined a distance function satisfying no specific conditions an abstract metric space. Birkhoff® has investigated spaces in which condition 2 is replaced by the condition a = 6 implies p(a, 6) = 0. Spaces with a so-called weak metric have been studied by Ribeiro.? Here the distance function satisfies condition 3 and the condition that a = b implies p(a, b) = 0 but does not necessarily satisfy conditions 1 and 2 (for example, the distance in a mountainous countryside might be measured by the time required to traverse it). Furthermore, spaces with the conditions 1, 2, and a weaker form of 3 have also been studied as, for example, the almost-metric spaces of Menger.}° Menger" has investigated spaces in which the distance is not necessarily a real number. Finally, is defined neighbourhood

Bieberbach”

calls

in it in such which

a (V)space

a manner

is a metric

that

space. 98

locally

metric

each

of the

if a distance elements

function possesses

a

Metric

SPACES

99

Examples of metric spaces 1. Let M be a given set and p(x, y) an arbitrary symmetric function of the two variables x and y defined for « € M, y € M in such a manner that p(x, y) = 0 for x = y, and 1 < p(w, y) 1+1=2>

p(x,2).

In particular, we may assume p(x, y) = 1 for x # y. Thus every set becomes a metric space if we assume that the distance between any two different elements is equal to 1. There exist, therefore, metric spaces of arbitrary power.

2. The set P of all points in the plane becomes a metric space if the distance p(pi, p2) between the points pi = (x1, yi) and pe-= (x2, ye) is defined to be the number

p(P1,P2) = ((%1—x2)”+ (91—ye2)”)}, where the radical is to be taken with the positive sign. But the set P will also become a metric space if the distance between the points 1 = (x1, y1) and po = (x2, ye) is defined to be the number

pi(Pi, Po) = |x1 — X2| + [x1 —yo! (since the function pi, as well as the function p, satisfies the three distance axioms). It is easily seen that we do not have p(f1, p2) = pi(f1, p2) for every pair pi, P2 of elements of P; there are, therefore, two different metrics p and pi defined in the set P. There arises now the question—how many different metrics can be defined in a given set? Consider

first

defined

as soon

distance

may

can

now

for

consisting

as the distance be any

be defined

readily

a set

real

finite

positive set

elements.

between

in a set consisting

every

of two

the two

number,

elements

it is clear

of two

containing

A metric

elements. more

than

in this

is given.

that The one

Since

c different same

set

is

this

metrics

result

element.

follows We

shall

prove

THEOREM 62. The set of all different metrics that can be established in an infinite set of cardinal m has cardinal 2™. Proof. of c™

two

A metric variables

= 2™ such

cardinal

in aset which

functions,

M of cardinal are

elements

there

are

m > No is defined of

at

the

most

m.

231095

set

M

and,

2™ different

by a real since metrics

function there

are

in a set of

100

Now

MeErtTrRIC SPACES

let £ denote

any

subset

of

M and

pz(a,b)

put

= 1

pze(a,b) =0

fora

€ Eandd

€ £,

ifa =b,

pz (a, b) =2 otherwise. The function pz(a, b) obviously satisfies the distance axioms (see example 1). If now £; and £; be any two different subsets of 1, each consisting of more than one element, and if a € FE; — Eo, 6 € Ai, or a€ Ep — FE,, b € Es, then pz,(a, 6) ¥ pz,(a, 6). Hence to different subsets of M, consisting of more than one element, correspond different metrics; since the set of these subsets has cardinal 2™, the set of all different metrics in the set M has cardinal > 2™. But, as proved above, there can be at most 2™ different metrics in the set M; hence the cardinal of the set of all different metrics is 2™. This proves the theorem.

47. Congruence of sets. Equivalence by division. Let E and F; be subsets of metric spaces M and M, respectively with metrics p in M and p; in M,. (In particular, we may have p = p; and even M = Mj.) If it is possible to establish a (1,1) correspondence f between the elements of E and £; in sucha manner that

(1)

pil(f(P),f(g) = eld, q)s

PE E and q Ez

then the set £ is said to be congruent to the set FE). The set FE; is then congruent to the set E since (1) is equivalent to the relation

o(f (a), f-(b)) = pi(a,b),

a@ € E,andb

€ Ey.

The sets Z and £, are then said to be congruent to each other or isometric; in symbols E~ E,. The relation of congruence of sets is therefore symmetric. Furthermore, it is transitive and reflexive.

A function f satisfying relation (1) is said to establish an isometric mapping of the set E on the set E, = f(£). It follows readily from (1) that f is (1,1) in FE. For if p#q, p € Eand g € E, we have p(p, g) ¥ 0, by condition 2. Hence, by (1), pi(f(), f() #0 and so, since p; satisfies condition 2, fp) #f(@)Relation (1) may be satisfied by two (or more) different functions f defined in Z. Thus two sets may be congruent in different ways (for example, two line segments of equal length are congruent either by translation or by rotation). In some cases, however, the congruence of two sets may be unique (e.g., the two half rays x < 0 and x > 0). Two metric spaces consisting of the same elements but with different metrics may be isometric. For example, the space consisting of the three elements 1, f2, and p3 with the metric p, where p(1, g2) = 3, p(p1, ps) = 4,

EQUIVALENCE BY DIVISION

101

p(p2, ps) = 5, and the space consisting of the same elements with metric pu, where pi(fi, p2) = 5, pilhi, ps) = 3 and pi(ps, p3) = 4. For, putting f(pi) = bs, f(b2) = pi, and f(p3) = p2, we obtain an isomorphic mapping f of the first space on the second. Certain theorems of an apparently simple nature concerning the congruence of sets are actually very difficult to establish, even if we restrict ourselves to the case of linear point sets, and have been proved only on the assumption of the axiom of choice. One such theorem is the theorem of Kuratowski!? which states: If a set E contained in a metric space be divided in two ways into the sum of two disjoint and congruent sets, say EH= M,+ M,= N,+ Nz, where MM,~ M,and N;, ~ Nz, then the sets WM,and JN, are each sums of four disjoint sets correspondingly congruent, that is M, = P; + P,+ P3+ Pi, Ni = Qi + Q2 + Os + Qu, where P, ~ Q, for i = 1, 2, 3, 4. The generalization of this theorem by K6nig and Valko! is even more difficult. This states: If a set E be divided in two ways into a finite number nm of disjoint and congruent subsets, say HF=M,+M.+...+ M, =Ni,+No+...+Nn, where M,;~ M,, Nim NN;for i= 1, 2,..., n, k =1,2,...,m, then the sets M,; and J, are each sums of n? disjoint sets: Mi=PitPet...+ Pa, Ni=QitQe+...+ Qn, where P:~O, fort = 1,2,...,n*. The proof is very difficult even for x = 3. A set may

be congruent

line x > 0 and its proper

subsets

the

half-line is called

to a proper «>

1.

subset

of itself;

A set which

monomorphic

for example

is not

congruent

the halfto any

of

by Lindenbaum.!®

It can be shown that every bounded linear set is monomorphic. It is easy, however, to construct a bounded set of points in the plane which is not monomorphic. For example, starting from a fixed point on the circumference of a circle, measure off along the circumference an infinite number of times an arc of fixed length which is not a rational multiple of the circumference, proceeding always in the same direction. The end-points of these arcs form the required set. One can also construct an unbounded plane set which is the sum of two disjoint sets each of which is congruent to the original set.‘° This set is obtained as follows: Let ¢ denote a translation of the plane through unit distance in the direction of the x-axis and W a rotation of the plane through one radian about the point (0, 0). Let E be the set obtained by the application of the two transformations ¢ and y to the point (0, 0) a finite number of times in any order.!”7 mnt A = ¢(£) and B=y(£). It is obvious that A~E, BYE, ACE, BCE,E=A+48B. Further, it can be shown that A.B = 0 (the proof rests on the transcendental nature of the radian). Furthermore,!* for every cardinal number m < Cc there exists a plane set P which is the sum of m disjoint sets each congruent to P. It can be shown,

102

Merric

SPACES

however, that no linear set can be expressed as the sum of two disjoint sets each congruent to the original set. Nor can a bounded plane set be so expressed. However, such sets exist in R; on the surface of a sphere. With the aid of the axiom of choice it can be proved that every segment of a straight line is the sum of a countable aggregate of disjoint sets all congruent to each other;!° and that, for every cardinalm < c¢, the straight line is the sum of m disjoint

sets all congruent

to each

other.”°

It is readily deduced that if a metric space of itself it is itself a proper subset of a metric and

is congruent to a proper subset space to which it is congruent

conversely.

If a metric space E is isometric with a subset of a metric space M, then we say that E can be embedded (metrically) in the space M. Clearly every metric space consisting of two elements can be embedded in a linear space and one consisting of three elements can be embedded in a plane (with the usual definition of distance). There exist, however, metric spaces consisting of four elements which cannot be embedded in a three-dimensional space (and not even in an v-dimensional space, as we shall show in § 57). If which

is any every

family metric

of metric space

spaces,

M of the

then

family

there @ can

exists

a metric

space

K in

be embedded.

For let K denote the set of all pairs (pb, M), where p € M € ®. Let py bea fixed element of the space M for each M € ®. Define the distance p(q:, ¢2) of two elements of K as follows:

If Ci

(pi, M), Op (pa,M2)and M, = M, € M, put p(Q1,G2) = pa (Pr, 2);

where py is the distance in the space M; if Mi # Mo, put p(Q1,92) = pau,(Pi, bu)

+ pa, (b2, Pa.) + 1.

It is easy to see that the function p so defined satisfies the three distance axioms and that the space M (M € ®) is congruent to the subset of K consisting of all the elements (p, 4), where » € M. Thus K has the required properties.

We note further that from the so-called Cantor hypothesis about the it follows that, for every cardinal numberm > No, there exists a space U,, of cardinalm in which every metric space of cardinalm embedded.”!_ In particular, the continuum hypothesis is equivalent existence of such a space Us

Two metric correspondence every point in sponding point

spaces are said to be between their points one space is isometric in the other.?* Two

alephs metric can be to the

locally congruent if there exists a (1, 1) under which some neighbourhood”? of with some neighbourhood of the corremetric spaces may be locally isometric

EQUIVALENCE BY DIVISION

103

without being isometric. For example, the linear space R; and the space P consisting of all real numbers in which the distance p(x, y) = min (|x—y], 1). Examples 1. Give

an example

is congruent

of two linear

to a certain

subset

sets which

are not congruent

although

of the other.

Let A consist of all real numbers > 1 and B = A + {0}. congruent to a subset of A obtained from B by a translation length. But the set A is not congruent to the set B.

Another example is supplied by the countable linear sets and B, = A, + {0}. 2. A set if there any

exists

two

points

T contained

in a metric

a (1, 1) mapping

points

in

space

is called

of £ on T under

T is k times

the

distance

which

set of all rationals, are

Construct

(c) disjoint

be the

A;= {2,3,4,...} larger

the

than

distance the

a set E between

corresponding

each

Rare med P=—%,+7.+..., than 7.

to a set twice as large.

or the set of all numbers

2*, where

sets.

set

T which

is the

sum

of two

disjoint

sets

each

of

as T.

set of all natural

numbers

Construct sets

such

as large

of all odd

the set of all reals

a linear

is twice

T may set

which

The set B is through unit

in E£.

Rkis an integer,

(b)

k times between

(a) Give an example of a linear set congruent The

each

and

a linear of which

the set

numbers

and the

set of all even

T which

is at least

is the twice

two disjoint

sets may

be the

numbers. sum

of a countable

as large

aggregate

of

as T.

t) 3, 5th and [geri Lea! ody 2a, 2 ees Upereenee %.7,=0 for 1 1) can be embedded in an (” — 1)-dimensional Euclidean space. There exist, however, countable subsets of Hilbert space which cannot be embedded in any Euclidean space of a finite number of dimensions (for example, the set consisting of the elements #1, fo, .. . ,where, form = 1, 2,..., p, is an infinite sequence whose mth term is 1 while all others are zero).

No sphere S(p, r) of Hilbert space is compact. For let p = (x1, X2,...) denote a point of Hilbert space-and put p, = (x1, %2,...) Xn-1) Xn+ 7/2,

HILBERT

SPACE

Naeine--) for n = 1, 2;.... Hence p, distance between any two different points consequently, the set S(p, r) D P is not taining it). Hilbert space is, therefore, points. It will, however, be shown later with a certain compact subset of itself. We prove next subset of itself.

that

Hilbert

space

125

€S(p, r) form = 1, 2;... and the of the set P = {f1, po,...} is 7/24; compact (in any metric space connot locally compact at any of its that Hilbert space is homeomorphic

is isometric

with a certain

nowhere-dense

To this end let f(p) = (0, x1, x2,...) for pb = (x1, %2,.--.) € H. Clearly H and its subset E = f(H) are isometric. Moreover, the set £ is nowheredense on H. For let g = (yi, y2,...) denote any point of H and ¢ an arbitrary positive number. Let a ~ 0 bea number such that y; < a < y1 + €/2 and put go = (G, yo, y3,---)Then p(qgo,g) = a — y1 < ¢/2. Put r = min ( lal, ¢/2). Since p(go, g) < «/2 andr < «/2, we have S(qo, r) CS(qo, €/2) C S(q, €). Suppose that p = (x1, %2,...) € S(qo, 7); then p(p, qo) 0 be a given n>~ number. Since p € H the series x)? + x,? +... is convergent; there exists, therefore, a natural number uwsuch that 1/p < e? and x41? + Xn? +... < form

> yw. But

nN

< 1/n,

Hiedeat Oe a

126

Metric

hence,

SPACES

for 7 > uz,

e(P, Pa) => (:,i =i) + Ssat) uy.

Consequently, limf(6™) = f(2). Supposethat p™is a sequenceofelementsofE, suchthat AeiG)(p™)= f(p) in H. From (27) and the properties of H, we obtain (28). Butifa/(1 + lal) == 6, then |b] < 1 and a = b/(1 — |d|). Hence if ai, a2,... is a sequence of real numbers such that

lim dn/(1 + |a@n|)= a/(1 + |a|), then lim a, = a. Therefore

(28) gives

lim x, n>co

In #,, this implies that

= x,

OE

lim p” = p. N-co Thus for p™ € E, and p € E, the relations lim p” = p(in E,) and lim f(p™) = f(p) (an H) are equivalent. Denote ordinates. irrational

This proves that f maps E, topologically on f(£,)

by H,, (Fréchet) We shall prove numbers.

= Q.

the set of all elements of E, with irrational cothat H, is homeomorphic with the set H; of all

For x real, put

o(x%)= 3+

3x/(1 + |x|);

the function ¢(«) obviously establishes a homeomorphism

between the set of

140

METRIC

all real numbers and = (2y — 1)/(1 -—2y set H, of all irrational interval (0, 1). Now

SPACES

the open interval (0, 1) (its inverse function is x(y) _ 1|) for 0 < y < 1) and therefore also between the numbers and the set 7) of all irrational numbers in the if, for p = (%1,%2,...) € Hu, we put

f(t) cz ((x1), (x2), oe)s we obtain a homeomorphic mapping of the elements of E, with irrational coordinates sufficient, therefore to show that Th 7). Let p = (x1, Xo,...) be a given element therefore irrational and in the interval (0,

set H, on the set T = f(H.) of all in the interval (0, 1). It will be

of J. The numbers 4%,%2,... 1); let

are

1 1 x, = n EeSee Gan Er wineo theese be the development of the number x; as a continued fraction. | Employ the diagonal method to rearrange the double sequence 1; into a single sequence M1, Ms, Ms3,.... Put 1 1 1 f@®) = ere mM,+ Me+ mM3+ It is easily shown that Th T,. It is sufficient here to base the proof on the properties of sequences converging to a given limit in £, and on the following two known properties of infinite continued fractions: (i) For every irrational number x and every positive integer k, there exists a positive number e such that every irrational number x, which satisfies the inequality |x — xo| < possesses a development as a continued fraction which is identical in the first k convergents with that of the number xp itself. (ii) For every irrational number x» and every positive number e, there exists a positive integer k such that every irrational number x; whose development into a continued fraction has the first k convergents identical with the corresponding convergents in the development of xo, satisfies the inequality |2c _ %a| =e The relation , h H; may, therefore, be considered proved. Translations of the space E,. Let ai, d2,... denote an infinite sequence of realnumbers. Associate with each element p = (x1, Xo,...) € E, the element (30)

o(p) = (x1 + ai, x2 + ae, ...).

It is easily seen that E, hy ¢(E.). The transformation ¢ is called a translation of EF, (by analogy with m-dimensional space). It follows readily from (30) that a translation of Z, is an isometric mapping of E, on itself. Since by a suitable translation any element of E, can be transformed into any other, it may be said that £, is not only topologically but also metrically homogeneous. Let now JN be any continuum. N maps

into

set of elements

We shall

show

a certain

subset

that

there

of H,.

of E,, with exists

cardinal

a translation

To this end we first

less than of E, prove

that under

of the which

the following

FRECHET’S

SPACE EF,

141

LemMA. If G ts a set of real numbers of cardinal < c, there exists a real number a such that for every number x of Q the number x + a is irrational. Proof. Let m be the cardinal of Q; then m2. Hence there exists no set E such that yE < yE2.

152

MeEtTrRIC SPACES

It follows that among the y-types of all linear sets there is none that is smaller than each of the others. For, if E were such a set, we would have vE < yE:, where E; = {0} + {1/n} for m = 1, 2,... which, from the above, would be impossible, since E; is closed and compact. A similar argument establishes the same result for the y-type of linear sets of cardinal c. However, among the y-types of all countable metric spaces there is one which is greater than that of each of the others. This is clearly the y-type of the set NV of all natural numbers. Sets of type yV are obviously isolated countable sets. Hence sets whose y-type is less than yNV must be countable sets containing at least one limit point. Among the y-types of these sets there is one which is greater than each of the others; namely the y-type of the set E; defined above. But there is no greatest y-type’! among all y-types which are less than yF. It has been shown that there exist non-countable transfinite sequences of increasing (decreasing) y-types of countable sets.7* There exists also a noncountable family of countable sets whose y-types are all incommensurable, i.e., no two can be related by one of the signs = , . There are probably N: different y-types of all countable linear sets but this has not so far been proved without the continuum hypothesis. It can be proved,’* however, that there are 2° different y-types of all linear sets of cardinal c. The

problem

of cardinal

whether

c, a linear

there

set whose

set of R is not as yet solved. R and

in some

cases

with

great

exists y-type

for every is greater

It is solved

family

R of ¢ linear

or equal

positively

only

sets,

to the y-type for certain

each

of each families

difficulties.”

Continuity-types. Two sets E; and E, contained in a metric space are said to have the same c-type (continuity) if each is a continuous image of the other.”® If Z; is a continuous image of EZ»but not conversely then we write cE, 0 corresponds

(53)

a number

p(p,g)

n > 0 such that the relation

0 such that pi(f(«), f(y)) < ¢ for x € E, y € E, and p(x, y) < 6.

UNIFORM CONVERGENCEOF SEQUENCES it x, 6 E.and

y, € E for n = 1, 2;...,

and lim’ p(x,, 4.) =,0, N30

155 then

there

exists an index yusuch that p(xn, Yn) < 6 for m > wand so pi(f (xn), f(vn)) < € form > yw. Since « is arbitrary this gives

Thus

the

condition

is necessary.

Suppose now that f is not uniformly continuous in E. Then there exists a number e > 0 such that for every natural m there exist points x, € E, y, € E for which p(x,, Yn) < 1/mnbut pi(f (xn), f(on)) > €. This gives

lim p(%a,Jn) = 90 but n-00 Hence

the condition

lim pi(f(xn), fOn)) # 0. N30

is sufficient.

64. Uniform convergence of a sequence of functions. Let E denote a given set contained in a metric space M with distance function p; let f,(p) (n = 1, 2,...) be an infinite sequence of functions defined in E and taking on values in a metric space M, with distance function p;. The sequence {fn(p)} is said to converge uniformly in E to the function f(p) defined in the set E and taking on values in M, if, for every « > Oand € E, there exists an index »(depending on ¢ebut not on ~) such that pi(f,(p), f(p)) < «form > p.

THEOREM 81. The limit of an infinite uniformly convergent sequence of continuous functions in a set E contained in a metric space M and taking on values im a metric space M, 1s continuous in E. Proof. Let {f,(p)} denote an infinite sequence of functions continuous in E and converging uniformly in that set to the function f(p). Let fo be an element of E and e an arbitrary positive number. Since the given sequence converges uniformly in £ there exists an index m such that

(62)

pi(fa(b),

(fP)) < €/3

forp

€ E

and so, in particular, (63)

p1(fn(bo), f(o))

Since f,(p) such that

(64)

is continuous

in E there exists,

< €/3. for the given e, a number

6 > 0

pi(fn(P), fn(bo)) < €/3 for p € Eand p(P, po) < 6.

Relations (62), (63), (64), and the triangle law give (65)

pi(f(e),

Thus the

for every

continuity

f(bo)) < efor p € Eand p(p, po) < 6.

¢ > 0 there

of the

function

exists

a 6 > 0 such

f in £.

that

(65) is true;

this

proves

156

Metric

SPACES

THEOREM82.89 An infinite sequence {fp(p)} of functions continuous im a closed and compact set contained in a metric space M and taking on values in a metric space M, converges uniformly in that set to a function f(p) uf and only af the condition

(66) impliesthe relation

lim pan= Po, sa

(67)

’, € E,n = 0, 1,2, see

limfn(Pn) = f (po).

Proof. Suppose that the sequence {f,(p)} of converges uniformly in E to the function f(p). denote an infinite sequence satisfying (66). Let number. Since, by Theorem 81, f is continuous in and so there exists an index u such that

(68)

functions continuous in E Let pz, m= 0, 1, 2,. e be an arbitrary positive E we have lim im f(r) = f (po)

pi(f (Pn) f(bo)) < €/2

where p; is the distance uniformly to f(p) in E;

form > H,

function in M;. But the sequence f,() hence there exists an index v such that

pilfn(h),

f(b))

< €/2

forp

converges

€ Eandnu

> vy

and so, in particular,

(69)

p1(fn(Pa)s f(Pn)) < e/2

forn > v.

Now (68), (69), and the triangle law give (70)

p1(fn(Pn), f(bo))
u+».

Thus corresponding to every positive ¢ there exists an index » + v such that (70) holds; this gives (67) and with it the necessity of the condition follows. Next suppose that £ is a closed and compact set and f,(p), 7 = 1,2,..., an infinite sequence of functions continuous in E for which condition (66) always implies (67). We first show that the function f is continuous in E. Suppose that f is not continuous in E at fo. Then there exists a number e > 0 and

an infinite

(71)

sequence

q;,, k = 1, 2,...,

in qx = poand pi(f(qe), f(bo))

The sequence p, = q, form = 1,2,...and

hence

of elements

for every

natural

exists

that

fork =1,2,....

the fact that (66) implies (67) give

lim fn(qu) = f(x); n-300

k there

>e

of E such

an index

J, > k such

k=

1, 25.

k=

Ly2; oeee

that

pil fre (Qx)sf(qe)) < €/2 and so, from (71), (72)

01( fu (Ge), f(Po)) a e/2,

UNIFORM CONVERGENCEOF SEQUENCES Since

], > k, lim ], = + K-00

© ; we may

therefore

remove

157 from

the

sequence

h,(k = 1,2,...) an infinite increasing sequence J,,(¢ = 1,2,...). Put pr = de: for m = |,; and p, = po for m not a term of the sequence i,;(4 = 1, 2,...). It follows from (71) that lim p, = po and so (66) holds; from (72), we get

pilfins(Pui),f(bo))a e/2 which contradicts (67). The function f is therefore continuous in E. Next, assume that the sequence {f,(p)} does not converge uniformly to the function f(p) in the closed and compact set E. Then there exists a positive number e¢such that to every natural / corresponds an index k; >/ and a point g; € & for which

(73)

GLC

DIAC),

Since £ is closed and compact there of indices /,(¢ = 1, 2, ...) such that

> &

exists an infinite

l= increasing

1; 2, . sequence

lim qi; = po € E, t4co whether the terms of the sequence {q;} are different or not. Furthermore, since f is continuous in £, there exists an index pwsuch that

pi(f(qi:)s and so, from (73), (74)

f(Po)) < €/2 -

pil frrs (qi), f(po)) > e/2

fori > u, fori > MK.

Since k,; > 1,;we have limk;, = + © ; thus there exists an infinite increasing 4-300 ' subsequence &;,,(s = 1, 2,...) of the sequence {k,,}. Put , = q:,. for n = k,,, and p, = po for m not a term of the sequence k;,,(s = 1, 2,...). Then ?%,;, = 41;, for s = 1, 2,.. and lim p, = po. Thus (66) is satisfied; from (74), ee pil ferse(Purse), F(Po)) > €/2 forz > uy, contrary

to (67).

This

proves

the theorem.

It is necessary that the set E in Theorem 82 be closed and compact. For suppose that E£ is either not closed or not compact. In either case there exists an infinite sequence 0 = {q1, qo, ...} of different points of E such that Q’.E=0. Put, forp € £,f,(p) = Oif p(P, dn) > 1/n, fr(b) = 1 — np(p, qn) if p(p, dn) < 1/n; also set f(p) = 0 for all p € £. It follows readily that the functions f,(p), m= 1, 2,..., are continuous in the set E. If fo is any point of E then, since Q’. E = 0, there exists a positive number 6 such that

(75)

p(p, bo) < 5, p € E, implies that p ¢ Q.

158

MerrRIC

Thus

if lim p, = po, Pp»€ E,

SPACES

n = 1, 2,...,

then,

for sufficiently

large nm,

S(Pny 1/n) C S(ho, 8). If p(Par Gn)